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electrical_engineering_and_electronics_2:block01 [2026/03/05 01:43] – angelegt mexleadminelectrical_engineering_and_electronics_2:block01 [2026/03/05 02:45] (aktuell) – mexleadmin
Zeile 3: Zeile 3:
 ===== Learning Objectives ===== ===== Learning Objectives =====
 <callout> <callout>
-After this 90-minute block, you can +After this 90-minute block, you  
-  * ...+  * know the time constant $\tau$ and in particularly calculate it. 
 +  * determine the time characteristic of the currents and voltages at the RC element for a given resistance and capacitance. 
 +  * know the continuity conditions of electrical quantities. 
 +  * know when (=according to which measure) the capacitor is considered to be fully charged/discharged, i.e. a steady state can be considered to have been reached. 
 +  * can calculate the energy content in a capacitor. 
 +  * can calculate the change in energy of a capacitor resulting from a change in voltage between the capacitor terminals. 
 +  * can calculate (initial) current, (final) voltage, and charge when balancing the charge of several capacitors (also via resistors).
 </callout> </callout>
 +
  
 ===== Preparation at Home ===== ===== Preparation at Home =====
Zeile 82: Zeile 89:
 </callout> </callout>
  
-===== 5.1 Time Course of the Charging and Discharging Process =====+====Time Course of the Charging and Discharging Process ====
  
-<callout> 
- 
-=== Learning Objectives === 
- 
-By the end of this section, you will be able to: 
-  - know the time constant $\tau$ and in particularly calculate it. 
-  - determine the time characteristic of the currents and voltages at the RC element for a given resistance and capacitance. 
-  - know the continuity conditions of electrical quantities. 
-  - know when (=according to which measure) the capacitor is considered to be fully charged/discharged, i.e. a steady state can be considered to have been reached. 
- 
-</callout> 
  
 In the simulation below you can see the circuit mentioned above in a slightly modified form: In the simulation below you can see the circuit mentioned above in a slightly modified form:
Zeile 113: Zeile 109:
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
- 
-Here is a short introduction about the transient behavior of an RC element (starting at 15:07 until 24:55) 
-{{youtube>8nyNamrWcyE?start=907&stop=1495}} 
  
 To understand the charging process of a capacitor, an initially uncharged capacitor with capacitance $C$ is to be charged by a DC voltage source $U_{\rm s}$ via a resistor $R$. To understand the charging process of a capacitor, an initially uncharged capacitor with capacitance $C$ is to be charged by a DC voltage source $U_{\rm s}$ via a resistor $R$.
Zeile 134: Zeile 127:
 C = {{q(t)}  \over{u_C(t)}} = {{{\rm d}q}  \over{{\rm d}u_C}} = {\rm const.} \tag{5.1.1}  C = {{q(t)}  \over{u_C(t)}} = {{{\rm d}q}  \over{{\rm d}u_C}} = {\rm const.} \tag{5.1.1} 
 \end{align*} \end{align*}
- 
-The following explanations are also well explained in these two videos on [[https://www.youtube.com/watch?v=csFh588BODY&ab_channel=MattAnderson|charging]] and [[https://www.youtube.com/watch?v=eCOLkUPSpxc&ab_channel=lasseviren1|discharging]]. 
- 
  
 ==== Charging a capacitor at time t=0 ==== ==== Charging a capacitor at time t=0 ====
Zeile 351: Zeile 341:
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-===== 5.2 Energy stored in a Capacitor ===== +==== Energy stored in a Capacitor ====
- +
-<callout> +
- +
-=== Learning Objectives === +
- +
-By the end of this section, you will be able to: +
-  - calculate the energy content in a capacitor. +
-  - calculate the change in energy of a capacitor resulting from a change in voltage between the capacitor terminals. +
-  - calculate (initial) current, (final) voltage, and charge when balancing the charge of several capacitors (also via resistors). +
- +
-</callout>+
  
 <WRAP> <imgcaption imageNo02 | circuit for viewing the charge curve> </imgcaption> {{drawio>SchaltungEntladekurve2.svg}} </WRAP> <WRAP> <imgcaption imageNo02 | circuit for viewing the charge curve> </imgcaption> {{drawio>SchaltungEntladekurve2.svg}} </WRAP>
  
-Now the capacitor as energy storage is to be looked at more closely. This derivation is also explained in [[https://www.youtube.com/watch?v=PTyB6_Kt_5A&ab_channel=TheOrganicChemistryTutor|this youtube video]]. For this, we consider again the circuit in <imgref imageNo02 > an. According to the chapter [[:electrical_engineering_1:preparation_properties_proportions#power_and_efficiency|Preparation, Properties, and Proportions]], the power for constant values (DC) is defined as:+Now the capacitor as energy storage is to be looked at more closely. For this, we consider again the circuit in <imgref imageNo02 > an. According to the chapter [[:electrical_engineering_1:preparation_properties_proportions#power_and_efficiency|Preparation, Properties, and Proportions]], the power for constant values (DC) is defined as:
  
 \begin{align*} \begin{align*}
Zeile 495: Zeile 474:
 ==== Worked examples ==== ==== Worked examples ====
  
-...+ 
 +<panel type="info" title="Exercise 5.2.1 Capacitor charging/discharging practice Exercise ">  
 +<WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> 
 + 
 +{{youtube>b_S-Ni5n-2s}} 
 + 
 +</WRAP></WRAP></panel> 
 + 
 +#@TaskTitle_HTML@# 5.2.2 Capacitor charging/discharging #@TaskText_HTML@# 
 + 
 +The following circuit shows a charging/discharging circuit for a capacitor. 
 + 
 +The values of the components shall be the following: 
 +  * $R_1 = 1.0 \rm k\Omega$ 
 +  * $R_2 = 2.0 \rm k\Omega$ 
 +  * $R_3 = 3.0 \rm k\Omega$ 
 +  * $C   = 1 \rm \mu F$ 
 +  * $S_1$ and $S_2$ are opened in the beginning (open-circuit) 
 + 
 +{{drawio>electrical_engineering_1:Exercise522setup.svg}} 
 + 
 +1. For the first tasks, the switch $S_1$ gets closed at $t=t_0 = 0s$. \\ 
 + 
 +1.1 What is the value of the time constant $\tau_1$? 
 + 
 +#@HiddenBegin_HTML~Solution1,Solution~@# 
 + 
 +The time constant $\tau$ is generally given as: $\tau= R\cdot C$. \\ 
 +Now, we try to determine which $R$ and $C$ must be used here. \\ 
 +To find this out, we have to look at the circuit when $S_1$ gets closed. 
 + 
 +{{drawio>electrical_engineering_1:Exercise522sol1.svg}} 
 + 
 +We see that for the time constant, we need to use $R=R_1 + R_2$. 
 + 
 +#@HiddenEnd_HTML~Solution1,Solution ~@# 
 + 
 +#@HiddenBegin_HTML~Result1,Result~@# 
 +\begin{align*} 
 +\tau_1 &= R\cdot C \\ 
 +       &= (R_1 + R_2) \cdot C \\ 
 +       &= 3~\rm ms \\ 
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~Result1,Result~@# 
 + 
 +1.2 What is the formula for the voltage $u_{R2}$ over the resistor $R_2$? Derive a general formula without using component values! 
 + 
 +#@HiddenBegin_HTML~Solution2,Solution~@# 
 + 
 +To get a general formula, we again look at the circuit, but this time with the voltage arrows. 
 + 
 +{{drawio>electrical_engineering_1:Exercise522sol2.svg}} 
 + 
 +We see, that: $U_1 = u_C + u_{R2}$ and there is only one current in the loop: $i = i_C = i_{R2}$\\ 
 +The current is generally given with the exponential function: $i_c = {{U}\over{R}}\cdot e^{-t/\tau}$, with $R$ given here as $R = R_1 + R_2$. 
 +Therefore, $u_{R2}$ can be written as: 
 + 
 +\begin{align*} 
 +u_{R2} &= R_2 \cdot i_{R2} \\ 
 +       &= U_1 \cdot {{R_2}\over{R_1 + R_2}} \cdot e^{-t/ \tau}  
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~Solution2,Solution ~@# 
 + 
 +#@HiddenBegin_HTML~Result2,Result~@# 
 +\begin{align*} 
 +u_{R2} = U_1 \cdot {{R_2}\over{R_1 + R_2}} \cdot e^{t/ \tau} 
 +\end{align*} 
 +#@HiddenEnd_HTML~Result2,Result~@# 
 + 
 +2. At a distinct time $t_1$, the voltage $u_C$ is charged up to $4/5 \cdot U_1$. 
 +At this point, the switch $S_1$ will be opened. \\ Calculate $t_1$! 
 + 
 +#@HiddenBegin_HTML~Solution3,Solution~@# 
 + 
 +We can derive $u_{C}$ based on the exponential function: $u_C(t) = U_1 \cdot (1-e^{-t/\tau})$. \\ 
 +Therefore, we get $t_1$ by: 
 + 
 +\begin{align*} 
 +u_C = 4/5 \cdot U_1              &=  U_1 \cdot (1-e^{-t/\tau}) \\ 
 +      4/5                        &=             1-e^{-t/\tau} \\ 
 +      e^{-t/\tau}                &=             1-4/5 = 1/5\\ 
 +         -t/\tau                 &=             \rm ln (1/5) \\ 
 +          t                      &= -\tau \cdot \rm ln (1/5) \\ 
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~Solution3,Solution ~@# 
 + 
 +#@HiddenBegin_HTML~Result3,Result~@# 
 +\begin{align*} 
 +          t                      &=  3~{\rm ms} \cdot 1.61 \approx 4.8~\rm ms \\ 
 +\end{align*} 
 +#@HiddenEnd_HTML~Result3,Result~@# 
 + 
 +3. The switch $S_2$ will get closed at the moment $t_2 = 10 ~\rm ms$. The values of the voltage sources are now: $U_1 = 5.0 ~\rm V$ and $U_2 = 10 ~\rm V$. 
 + 
 +3.1 What is the new time constant $\tau_2$? 
 + 
 +#@HiddenBegin_HTML~Solution4,Solution~@# 
 + 
 +Again, the time constant $\tau$ is given as: $\tau= R\cdot C$. \\ 
 +Again, we try to determine which $R$ and $C$ must be used here. \\ 
 +To find this out, we have to look at the circuit when $S_1$ is open and $S_2$ is closed. 
 + 
 +{{drawio>electrical_engineering_1:Exercise522sol4.svg}} 
 + 
 +We see that for the time constant, we now need to use $R=R_3 + R_2$. 
 + 
 +\begin{align*} 
 +\tau_2 &= R\cdot C \\ 
 +       &= (R_3 + R_2) \cdot C \\ 
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~Solution4,Solution ~@# 
 + 
 +#@HiddenBegin_HTML~Result4,Result~@# 
 +\begin{align*} 
 +\tau_2 &= 5~\rm ms \\ 
 +\end{align*} 
 +#@HiddenEnd_HTML~Result4,Result~@# 
 + 
 +3.2 Calculate the moment $t_3$ when $u_{R2}$ is smaller than $1/10 \cdot U_2$. 
 + 
 +#@HiddenBegin_HTML~Solution5,Solution~@# 
 + 
 +To calculate the moment $t_3$ when $u_{R2}$ is smaller than $1/10 \cdot U_2$, we first have to find out the value of $u_{R2}(t_2 = 10 ~\rm ms)$, when $S_2$ just got closed. \\ 
 +  * Starting from $t_2 = 10 ~\rm ms$, the voltage source $U_2$ charges up the capacitor $C$ further. 
 +  * Before at $t_1$, when $S_1$ got opened, the value of $u_c$ was: $u_c(t_1) = 4/5 \cdot U_1 = 4 ~\rm V$. 
 +  * This is also true for $t_2$, since between $t_1$ and $t_2$ the charge on $C$ does not change: $u_c(t_2) = 4 ~\rm V$. 
 +  * In the first moment after closing $S_2$ at $t_2$, the voltage drop on $R_3 + R_2$ is: $U_{R3+R2} = U_2 - u_c(t_2) = 6 ~\rm V$. 
 +  * So the voltage divider of $R_3 + R_2$ lead to $ \boldsymbol{u_{R2}(t_2 = 10 ~\rm ms)} =  {{R_2}\over{R_3 + R2}} \cdot U_{R3+R2} = {{2 {~\rm k\Omega}}\over{3 {~\rm k\Omega} + 2 {~\rm k\Omega} }} \cdot 6 ~\rm V =  \boldsymbol{2.4 ~\rm V} $ 
 + 
 +We see that the voltage on $R_2$ has to decrease from $2.4 ~\rm V $ to $1/10 \cdot U_2 = 1 ~\rm V$. \\ 
 +To calculate this, there are multiple ways. In the following, one shall be retraced: 
 +  * We know, that the current $i_C = i_{R2}$ subsides exponentially: $i_{R2}(t) = I_{R2~ 0} \cdot {\rm e}^{-t/\tau}$ 
 +  * So we can rearrange the task to focus on the change in current instead of the voltage. 
 +  * The exponential decay is true regardless of where it starts. 
 + 
 +So from ${{i_{R2}(t)}\over{I_{R2~ 0}}} =  {\rm e}^{-t/\tau}$ we get  
 +\begin{align*} 
 +{{i_{R2}(t_3)}\over{i_{R2}(t_2)}} &=                                 {\rm exp} \left( -{{t_3 - t_2}\over{\tau_2}}       \right) \\ 
 +-{{t_3 - t_2}\over{\tau_2}}       &=                                 {\rm ln } \left( {{i_{R2}(t_3)}\over{i_{R2}(t_2)}} \right) \\ 
 +   t_3                            &= t_2          - \tau_2     \cdot {\rm ln } \left( {{i_{R2}(t_3)}\over{i_{R2}(t_2)}} \right) \\ 
 +   t_3                            &= 10 ~{\rm ms} - 5~{\rm ms} \cdot {\rm ln } \left( {{1 ~\rm V   }\over{2.4 ~\rm V }} \right) \\ 
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~Solution5,Solution ~@# 
 + 
 +#@HiddenBegin_HTML~Result5,Result~@# 
 +\begin{align*} 
 +t_3 &= 14.4~\rm ms \\ 
 +\end{align*} 
 +#@HiddenEnd_HTML~Result5,Result~@# 
 + 
 +3.3 Draw the course of time of the voltage $u_C(t)$ over the capacitor. 
 + 
 +{{drawio>electrical_engineering_1:Exercise522task6.svg}} 
 + 
 + 
 +#@HiddenBegin_HTML~Result6,Result~@# 
 +{{drawio>electrical_engineering_1:Exercise522sol6.svg}} 
 +#@HiddenEnd_HTML~Result6,Result~@# 
 + 
 +#@TaskEnd_HTML@# 
 + 
 +{{page>aufgabe_7.2.6_mit_rechnung&nofooter}} 
 + 
 +#@TaskTitle_HTML@#5.2.4 Charge balance of two capacitors \\ <fs medium>(educational exercise, not part of an exam)</fs>#@TaskText_HTML@# 
 + 
 + 
 +<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=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-KmvXR-oW9gADyoXz05JI6b09RQ7SPVDA5LwfBIp40EHu2UKd+QKTfeln3wfpid4aABIRsAB8izgId7YJYeD3EQDgpEBsawTg5K0BgqDknw0G3seGAkCEOLVOSQFbLBKjtFhIhCPhHDHigWE4pAEBEaSlCUceuB8MwDxPHWID8c2mQgJwoycDkIwSUiACWgoQkygoDEySIABaMEyjDzn4cmMC0ILgrpsItDk8KIn66K9CARIQJsxLGGY6KAIhEJLZCmlLUrSCrMqyKotFyUq8gKwpihKKoymF8qSrQloqkyIZ6sqyoAA7bmaFpWllB5jh24QMOWSQCMZXqmeZllKi0AZJeaDLpTuJR7nuNZPJAz4fJeXTZJMjRthUbUCPWl6dj1GBVmUA2ugIBV3P0BXZC5kytROYRJs4a2kr1raTYWq0dF2NxbeNTRlMadBrMwayoKgKQYEh+jXIIsyhJ2-YSBw1y1LscQ9lYH3sIKpKbIeyBvmYGYOe0QZphiMAqHFgMMASlig+DegKIS0Nxcj5Lw-dkRA6+oho2xFjiIg0DsdimYEymnn44jRP2STYNk-9lMQOS5PwHT95w7AdNI10INsz2FOY-c-0I2SNmM5EJDzOi7GWAAkhCcnyaMTIUoKIxIkMfgTEAA noborder}} </WRAP> 
 + 
 +In the simulation, you see the two capacitors $C_1$ and $C_2$ (The two small resistors with $1 ~\rm µ\Omega$ have to be there for the simulation to run). At the beginning, $C_1$ is charged to $10~{\rm V}$ and $C_2$ to $0~{\rm V}$. With the switches $S_1$ and $S_2$ you can choose whether 
 + 
 +  - the capacitances $C_1$ and $C_2$ are shorted, or 
 +  - the capacitors $C_1$ and $C_2$ are connected via resistor $R$. 
 + 
 +On the right side of the simulation, there are some additional "measuring devices" to calculate the stored potential energy from the voltages across the capacitors. 
 + 
 +In the following, the charging and discharging of a capacitor are to be explained with this construction. 
 + 
 +~~PAGEBREAK~~ ~~CLEARFIX~~  
 + 
 +Under the electrical structure, the following quantities are shown over time: 
 + 
 +^Voltage $u_1(C_1)$ of the first capacitor^Voltage $u_2(C_2)$ of the second capacitor^Stored energy $w_1(C_1)$^Stored energy $w_2(C_2)$^Total energy $\sum w$| 
 +|Initially charged to $10~{\rm V}$|Initially neutrally charged ($0~{\rm V}$)|Initially holds: \\ $w_1(C_1)= {1 \over 2} \cdot C \cdot U^2 = {1 \over 2} \cdot 10~{\rm µF} \cdot (10~{\rm V})^2 = 500~{\rm µW}$ \\ In the oscilloscope, equals $1~{\rm V} \sim 1~{\rm W}$|Initially, $w_2(C_2)=0$ , since the capacitor is not charged.|The total energy is $w_1 + w_2 = w_1$| 
 + 
 +The capacitor $C_1$ has thus initially stored the full energy and via closing of the switch, $S_2$ one would expect a balancing of the voltages and an equal distribution of the energy $w_1 + w_2 = 500~\rm µW$. 
 + 
 +  - Close the switch $S_2$ (the toggle switch $S_1$ should point to the switch $S_2$). What do you find? 
 +      - What do the voltages $u_1$ and $u_2$ do? 
 +      - What are the energies and the total energy? \\ How is this understandable with the previous total energy? 
 +  - Open $S_2$ - the changeover switch $S_1$ should not be changed. What do you find? 
 +      - What do the voltages $u_1$ and $u_2$ do? 
 +      - What are the energies and the total energy? \\ How is this understandable with the previous total energy? 
 +  - Repeat 1. and 2. several times. Can anything be deduced regarding the distribution of energy? 
 +  - Change the switch $S_2$ to the resistor. What do you observe? 
 +      - What do the voltages $u_1$ and $u_2$ do? 
 +      - What are the energies and the total energy? \\ How is this understandable with the previous total energy? 
 + 
 +#@TaskEnd_HTML@# 
  
 ===== Embedded resources ===== ===== Embedded resources =====
 +<WRAP>
 <WRAP column half> <WRAP column half>
-Explanation (video): ...+Here is a short introduction about the transient behavior of an RC element (starting at 15:07 until 24:55) 
 +{{youtube>8nyNamrWcyE?start=907&stop=1495}}
 </WRAP> </WRAP>
  
-~~PAGEBREAK~~ ~~CLEARFIX~~+<WRAP column half> 
 +Mathematical explanation of charging a capacitor 
 +{{youtube>csFh588BODY}} 
 +</WRAP> 
 +</WRAP> 
 +\\ \\ \\ \\ 
 +<WRAP> 
 +<WRAP column half> 
 +Mathematical explanation of discharging a capacitor 
 +{{youtube>eCOLkUPSpxc}} 
 +</WRAP>
  
 +<WRAP column half>
 +Mathematical explanation of the energy stored in the capacitor
 +{{youtube>PTyB6_Kt_5A}}
 +</WRAP>
 +</WRAP>
  
 +~~PAGEBREAK~~ ~~CLEARFIX~~